The following is a question from chapter 8 (Cauchy's Theorem) of Ian Stewart and David Tall's textbook on complex analysis. This is not a homework problem, I am self-teaching.
Let $$\gamma_1(t)=-1+e^{it}/2$$ $$\gamma_2(t) = 1+e^{it}/2$$ $$\gamma(t)=2e^{-it}$$ where $t\in[0,2\pi]$ for each of the paths. Let $f(z)=1/(z^2-1)$. Use Cauchy's generalized integral theorem to deduce $$\int_\gamma f = \int_{\gamma_1} f + \int_{\gamma_2} f$$
My attempt: The theorem in question says that, if $f$ is differentiable on a domain $D$, and $\gamma_1,\cdots,\gamma_n$ are closed contours in $D$ such that $w(\gamma_1, z) + \cdots + w(\gamma_n, z) = 0$ for all $z\notin D$, we have $$\int_{\gamma_1} f + \cdots + \int_{\gamma_n} f = 0.$$ ($w$ denotes the winding number or index).
For this question, $f$ is differentiable on $\mathbb{C}\setminus\{-1, 1\}$, so $z=\pm 1$ are the points of interest. By my understanding, it would be very convenient if we could show that $$w(\gamma_1, \pm 1) + w(\gamma_2, \pm 1) + w(-\gamma, \pm 1) = 0.$$ But for both $1$ and $-1$, the LHS is $2$. On the contrary, $$w(\gamma_1, \pm 1) + w(\gamma_2, \pm 1) + w(\gamma, \pm 1)=0$$ is true, so surely $$\int_{\gamma_1} f + \int_{\gamma_2} f = -\int_\gamma f,$$ which can only be reconciled with the problem statement if the RHS integral is $0$ (which it is, by direct computation) but how can I get to the quoted equality? (That is, how can I just get to the quoted result without appealing to the fact that the integral is 0)